Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
\(\Leftrightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)
\(\Leftrightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)
+) Nếu \(x+y+z+t\ne0\)
\(\Leftrightarrow y+z+t=z+t+x=t+x+y=x+y+z\)
\(\Leftrightarrow x=y=z=t\ne0\)
Mà \(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
\(\Leftrightarrow P=\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}\)
\(\Leftrightarrow P=\dfrac{2x}{2x}+\dfrac{2x}{2x}+\dfrac{2x}{2x}+\dfrac{2x}{2x}\)
\(\Leftrightarrow P=4\)
+) Nếu \(x+y+z+t=0\)
\(\Leftrightarrow x+y=-\left(z+t\right)\)
\(\Leftrightarrow\dfrac{x+y}{z+t}=\dfrac{-\left(z+t\right)}{z+t}=-1\)
Tương tự ta có :
\(\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}=-1\)
\(\Leftrightarrow P=-4\)
Vậy ..
Ta có: \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{y+t+x}=\dfrac{t}{y+x+z}\)
\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{y+t+x}+1=\dfrac{t}{y+x+z}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{y+t+x}=\dfrac{x+y+z+t}{y+x+z}\)+) Xét \(x+y+z+t=0\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\end{matrix}\right.\)
\(\Rightarrow A=-1\)
+) Xét \(x+y+z+t\ne0\Rightarrow x=y=z=t\)
\(\Rightarrow A=1\)
Vậy A = -1 hoặc A = 1
Ta có:\(\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{y+t+x}+1=\dfrac{t}{y+x+z}+1\)\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)
Nếu x+y+z+t\(\ne\)0 thì y+z+t=z+t+x=t+x+y=x+y+z
=>x=y=z=t nên P=1+1+1+1=4
Nếu X+y+z+t=0 thì P=-4
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{1}{3}=\dfrac{x+y}{\left(x+y\right)+2\left(z+t\right)}\)
\(\Rightarrow\left(x+y\right)+2\left(z+t\right)=3\left(x+y\right)\)
\(\Rightarrow2\left(z+t\right)=2\left(x+y\right)\Rightarrow\dfrac{x+y}{z+t}=1\)
Chứng minh tương tự ta được:
\(\dfrac{y+z}{x+t}=1;\dfrac{z+t}{x+y}=1;\dfrac{t+x}{y+z}=1\)
\(\Rightarrow P=1+1+1+1=4\)
+Xét x+y+z+t=0
\(\Rightarrow\)\(\left\{{}\begin{matrix}z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\\x+y=-\left(z+t\right)\\y+z=-\left(t+x\right)\end{matrix}\right.\)
Khi đó M=-4
+Xét x+y+z+t\(\ne\)0
ADTC dãy tỉ số bằng nhau ta có
\(\dfrac{x}{y+z+t}\)=\(\dfrac{y}{x+y+t}\)=\(\dfrac{z}{x+y+t}\)=\(\dfrac{z}{x+y+t}\)=\(\dfrac{x+y+z+t}{3.\left(x+y+z+t\right)}\)=\(\dfrac{1}{3}\)
+Với\(\dfrac{x}{y+z+t}\)=\(\dfrac{1}{3}\)
\(\Rightarrow\)3x=y+z+t
\(\Rightarrow\)4x=x+y+z+t
Chứng minh tương tự ta có
4y=x+y+z+t
4z=x+y+z+t
4t=x+y+z+t
Do đó x=y=z=t
Khi đó M=4
Ta có: \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\) (*)
+) Nếu \(x+y+z+t\ne0\) thì từ (*) suy ra:
\(y+z+t=z+t+x=t+x+y=x+y+z\)
\(\Rightarrow x=y=z=t\)
\(\Rightarrow P=\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}\) \(\Rightarrow P=1+1+1+1=4\)
+) Nếu \(x+y+z+t=0\) thì \(\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(t+x\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\)\(\Rightarrow P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Vậy \(P=4\) hoặc \(P=-4\)
\(\dfrac{x}{y+z+t}=\dfrac{y}{x+t+z}=\dfrac{z}{t+y+x}\)
\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{x+t+z}+1=\dfrac{z}{t+y+x}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{x+t+z}=\dfrac{x+y+z+t}{t+y+x}\)
\(\Rightarrow\left[{}\begin{matrix}x=y=z=t\\x+y+z+t=0\end{matrix}\right.\)
\(\circledast\) Khi \(x=y=z=t\) thì
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{x+t}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}=1+1+1+1=4\)
\(\circledast\) Khi \(x+y+z+t=0\) thì:\(\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{x+t}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Ta có :
\(\dfrac{x}{y+z+t}=\dfrac{y}{x+z+t}=\dfrac{z}{x+y+t}=\dfrac{t}{x+y+z}\)\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{x+z+t}+1=\dfrac{z}{x+y+t}+1\)\(=\dfrac{t}{x+y+z}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{x+y+t}=\dfrac{x+y+z+t}{x+z+t}=\dfrac{x+y+z+t}{x+y+z}\)
\(=\dfrac{x+y+z+t}{x+y+z}\)
* Nếu \(x+y+z+t=0\)
\(\Rightarrow x+y=-\left(z+t\right)\)
\(y+z=-\left(t+x\right)\)
Thay vào A ta được: \(P=-1+-1=-2\)
*Nếu \(x+y+z+t\ne0\)
\(\Rightarrow x+y+t=x+y+z\Rightarrow t=z\)
Làm tương tự tự ta suy ra được \(x=y=z=t\)
=> \(x+y=z+t\)
\(y+z=t+x\)
Thay vào A ta được A= 1+1=2
Vậy... tik mik nha !!!
Xét:
\(\dfrac{x}{y+z+t}+1=\dfrac{y}{x+t+z}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)
\(\Leftrightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)
+ TH1: Nếu \(x+y+z+t\ne0\Rightarrow x=y=z=t\Rightarrow P=4\)
+ TH2: Nếu \(x+y+z+t=0\Rightarrow P=-4\)
Vậy \(\left[{}\begin{matrix}P=4\\P=-4\end{matrix}\right.\)
Theo dãy tỉ số = nhau ta có :
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3x+3y+3z+3t}=\dfrac{1}{3}\)
\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Leftrightarrow3x=y+z+t\) (1)
\(\dfrac{y}{z+t+x}=\dfrac{1}{3}\Leftrightarrow3y=z+t+x\) (2)
\(\dfrac{z}{t+x+y}=\dfrac{1}{3}\Leftrightarrow3z=t+x+y\) (3)
\(\dfrac{t}{x+y+z}=\dfrac{1}{3}\Leftrightarrow3t=x+y+z\) (4)
Từ (1) và (2) => 3x + 3y = x + y + 2(z+t) => 2(x+y) = 2(z+t) => x + y = z + t (5)
Từ (2) và (3) => 3y + 3z = y + z + 2(t + x) => 2(y+z) = 2(t+x) = > y + z = t + x
Vậy P = \(\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}=4\)
Từ \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)
Vì \(x+y+z+t\ne0\) nên ta đi xét \(x+y+z+t=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(t+x\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\). Khi đó
\(P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=4\)
Sai đề chỗ p/s cuối. Xét 2 t/h.
Oáp Z_z có gì mai ns nhé!
mk ko làm cụ thể nhưng chỉ nêu hướng lm thôi nhé
bn áp dụng tích chất dãy tỉ số bằng nhau vào giả thiết, ra 1/3
sau đó suy ra x = (y+z+t)/3, y,z,t cũng làm tương tự
sau đó bạn quy đồng các mẫu của P
sau khi phân tích bn sẽ lấy kq vừa tính đc phần trên
mk nghĩ kết quả ra là 15 nhưng có thể sai
chúc bn may mắn